Answer:
Approximately [tex]641\; {\rm N}[/tex] (assuming that [tex]g = (-9.81)\; {\rm N \cdot kg^{-1}}[/tex].)
Explanation:
Forces on this person:
Let [tex]m[/tex] denote the mass of this person. It is given that [tex]m = 50\; {\rm kg}[/tex]. The weight of this person will be:
[tex]\begin{aligned}(\text{weight}) &= m\, g \\ &= (50\; {\rm kg})\, ((-9.81)\; {\rm N\cdot kg^{-1}}) \\ &\approx 490.5\; {\rm N}\end{aligned}[/tex].
(The value of weight is negative since weight points downwards.)
Let [tex]a[/tex] denote the acceleration of this person. It is given that [tex]a = 3\; {\rm m\cdot s^{-2}}[/tex] (since this person is accelerating upwards, the value of [tex]a[/tex] will be positive.) The net force on this person will be:
[tex]\begin{aligned}(\text{net force}) &= (\text{mass})\, (\text{acceleration})\\ &= (50\; {\rm kg})\, (3\; {\rm m\cdot s^{-2}}) \\ &\approx 150\; {\rm N}\end{aligned}[/tex].
Note that this net force is also equal to the vector sum of all the forces on this person. In other words:
[tex]\begin{aligned}(\text{net force}) &= (\text{weight}) + (\text{normal force})\end{aligned}[/tex].
Rearrange this equation to find [tex](\text{normal force})[/tex]:
[tex]\begin{aligned}(\text{normal force}) &= (\text{net force}) - (\text{weight}) \\ &\approx 150\; {\rm N} - ((-490.5)\; {\rm N}) \\ &\approx 641\; {\rm N}\end{aligned}[/tex].
Hence, the upward force on this person from the floor of the elevator will be approximately [tex]641\; {\rm N}[/tex].
